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Natasha2012 [34]
2 years ago
9

8. How much heat, in kilojoules, is needed to produce 1850 grams of ammonia (NH3),

Chemistry
1 answer:
AleksAgata [21]2 years ago
6 0

Based on the enthalpy of the reaction, 41625 kJ of heat is needed to produce 1850 grams of ammonia (NH3).

<h3>What is heat of reaction?</h3>

The heat of reaction or enthalpy change is the heat absorbed or evolved when reactant molecules react to form products.

According the enthalpy value of the reaction, 4 moles of ammonia require 1530 kJ of heat for its formation.

Molar mass of ammonia = 17 g

Mass of 4 moles of ammonia = 4 × 17 = 68 g

1850 g of ammonia will require = 1850 × 1530/68 = 41625 kJ of heat.

Therefore, 41625 kJ of heat is needed to produce 1850 grams of ammonia (NH3).

Learn more about enthalpy at: brainly.com/question/14047927

#SPJ1

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Consider the following reaction:
adell [148]

Answer:

1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹

2. 0.58 mol

Explanation:

1.Given ΔO₂/Δt…

    2H₂O₂     ⟶      2H₂O     +     O₂

-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt  

d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹

 d[H₂O]/dt =  2d[O₂]/dt =  2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ =  6.6 × 10⁻³mol·L⁻¹s⁻¹

2. Moles of O₂  

(a) Initial moles of H₂O₂

\text{Moles} = \text{1.5 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{1.5 mol }

(b) Final moles of H₂O₂

The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.

\text{Moles} = \text{1.5 L} \times \dfrac{\text{0.22 mol}}{\text{1 L}} = \text{0.33 mol }

(c) Moles of H₂O₂ reacted

Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol

(d) Moles of O₂ formed

\text{Moles of O}_{2} = \text{1.33 mol H$_{2}$O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol H$_{2}$O}_{2}} = \textbf{0.58 mol O}_{2}\\\\\text{The amount of oxygen formed is $\large \boxed{\textbf{0.58 mol}}$}

8 0
3 years ago
What is the enthalpy of reaction?
erma4kov [3.2K]

Answer:

Enthalpy is the amount of heat released or absorb. The total enthalpy can be found by subtracting the products' enthalpy by the reactants' enthalpy

4 0
3 years ago
Read 2 more answers
The chart shows the solubility of different substances.
Annette [7]
Answer: 1) Temperature can change the solubility of a solute.

Explanation:

The chart is missing so there is no way to tell what does the graph show.

Yet, I can help you because I can explain the status of each statement of the choices. As you will see there is only one possibility..

<span>1) Temperature can change the solubility of a solute.

Yes, temperature definetly can, and mostly do, modify the solubility of a solute.

You can search any chart of solubility and will find that.

I can give you two examples:

a) Sodium chloride: dissolve some spoons of salt in a cold water  until you can not dissolve more. Then, heat the water, you will find that more salt will get dissolved, proving that the temperature of the solution increases the solubility of sodium chloride.

b) Carbon dioxide gas: the soft drinks have CO₂ molecules dissolved in it.
 
The higher the temperature of the soft drink the less the amount of CO₂(g) that can be dissolved. That is why the soda bottling plants cool the beverage before adding the CO₂(g).

2) </span><span>Temperature has no affect on the solubility of a solute.

Since this is the opposite to the first statement and the first is true, this is false.

3) Salt has a greater solubility than sugar.

False.

This is an empirical result, which you cannot predict theoretically. So you need to see at the data either in a table or in a chart. Else you can test it at home. After the empirical data are shown it results that more grams of sugar can be dissolved in water compared to salt.

That is something you ca see in a chart or you can prove by yourself.

4) Nitrite salt has a greater solubility than sugar. </span>

False.

Looking at some data you can find that sodium nitrite solutiliby is aroun  70 - 100 g/10 g while sugar (sucrose) solutiblity is around 180 - 235 g/ 100 g.

8 0
3 years ago
Read 2 more answers
The isomerization of cyclopropane to propylene is a first-order process with a half-life of 19 min at 500°c. the time it takes f
amid [387]
<span>As mentioned, the isomerization of cyclopropane to propylene is a first-order process with a half-life of 19 min at 500°c. A first-order reaction kinetic rates means that the rate is constant throughout the reaction.

Thus, the time it takes for the partial pressure of cyclopropane to decrease from 1 atm to 0.125 atm at 500°c is </span><span>57 minutes.</span>
3 0
3 years ago
Lead has a density of 11.4 g/cm^3. What is the density in kilograms per cubic meter?
Vikentia [17]

The density in kg/m³ = 1.14 x 10⁴

<h3>Further explanation  </h3>

Density is a quantity derived from the mass and volume  

Density is the ratio of mass per unit volume  

With the same mass, the volume of objects that have a high density will be smaller than objects with a smaller type of density  

The unit of density can be expressed in g/cm³ or kg/m³  

Density formula:  

\large {\boxed {\bold {\rho ~ = ~ \frac {m} {V}}}}

ρ = density , g/cm³ or kg/m³  

m = mass  , g or kg

v = volume  , cm³ or m³  

A density of Lead : ρ = 11.4 g/cm³

the density in kg/m³ :

\tt 11.4~\dfrac{g}{cm^3}\times \dfrac{kg}{10^3~g}\times \dfrac{cm^3}{10^{-6}~m^3}=\boxed{\bold{1.14\times 10^4~\dfrac{kg}{m^3}}}

8 0
3 years ago
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